Question

If \[ z=x^2y^3+e^y\sin x, \] then \[ \frac{\partial^2 z}{\partial x\partial y}= \]

• A. \(6xy^2+e^y\cos x\)
• B. \(3x^2y^2+e^y\sin x\)
• C. \(3x^2y^2+e^y\cos x\)
• D. \(6xy^2+e^y\sin x\)

Hint:

In partial derivatives, treat the other variable as constant while differentiating.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a problem in partial differentiation. We need to find the mixed second-order partial derivative of the function \( z \) with respect to \( y \) and then \( x \).
Partial differentiation with respect to one variable involves treating all other variables as constants.
Step 2: Key Formula or Approach:
1. First find \( \frac{\partial z}{\partial y} \).
2. Then differentiate the result with respect to \( x \).
Step 3: Detailed Explanation:

First Partial Derivative (\( \partial z / \partial y \)):
Given \( z = x^2 y^3 + e^y \sin x \).
Treat \( x \) as a constant:
\[ \frac{\partial z}{\partial y} = x^2 \frac{\partial}{\partial y}(y^3) + (\sin x) \frac{\partial}{\partial y}(e^y) \]
\[ \frac{\partial z}{\partial y} = x^2 (3y^2) + (\sin x) (e^y) = 3x^2 y^2 + e^y \sin x \]

Second Mixed Partial Derivative (\( \partial^2 z / \partial x \partial y \)):
Now, differentiate the above result with respect to \( x \), treating \( y \) as a constant:
\[ \frac{\partial}{\partial x} \left(3x^2 y^2 + e^y \sin x\right) = 3y^2 \frac{\partial}{\partial x}(x^2) + e^y \frac{\partial}{\partial x}(\sin x) \]
\[ = 3y^2 (2x) + e^y (\cos x) \]
\[ = 6xy^2 + e^y \cos x \]

Step 4: Final Answer:
The mixed partial derivative is \( 6xy^2 + e^y \cos x \).